Rational Exponents: Additional Practice (Set 1)

The negative exponent flips the fraction, and then the cube root acts on the numerator and denominator separately.

$\left(\frac{8}{125}\right)^{-2/3} = \left(\frac{125}{8}\right)^{2/3} = \left(\sqrt[3]{\frac{125}{8}}\right)^{2} = \left(\frac{5}{2}\right)^{2} = \frac{25}{4}$

Flipping first keeps every exponent positive, which makes the arithmetic straightforward.

Inside the bracket, dividing subtracts the exponents, and subtracting a negative adds.

$\frac{x^{3/4}}{x^{-1/2}} = x^{\frac34 + \frac12} = x^{\frac34 + \frac24} = x^{5/4}$

Now raise to the eighth power, multiplying the exponent by $8$.

$\left(x^{5/4}\right)^{8} = x^{10}$

Convert each choice to a single rational exponent. The two routes agree for a positive base, so $\sqrt[6]{x^{5}}$ and $\left(\sqrt[6]{x}\right)^{5}$ are both $x^{5/6}$, and the product adds exponents: $\tfrac12 + \tfrac13 = \tfrac56$.

$\sqrt[5]{x^{6}} = x^{6/5} \neq x^{5/6}$

The odd one out swaps the root index with the power, turning $\tfrac56$ upside down.

Take the three powers one at a time, remembering that any nonzero base to the zero power is $1$.

$16^{1/2} = 4, \qquad 16^{1/4} = 2, \qquad 16^{0} = 1$

Now add.

$4 + 2 + 1 = 7$

Each exponent is a root, not a divisor, so the terms shrink as the denominator grows.

All three are positive, and raising positive numbers to a positive power preserves their order. Raise everything to the sixth power, which clears both denominators.

$\left(5^{1/2}\right)^{6} = 5^{3} = 125, \qquad 2^{6} = 64, \qquad \left(10^{1/3}\right)^{6} = 10^{2} = 100$

Since $64 < 100 < 125$, the original order is $2 < 10^{1/3} < 5^{1/2}$.

Resolve each power of a power first, multiplying exponents.

$\left(a^{1/3}\right)^{2} = a^{2/3}, \qquad \left(a^{1/2}\right)^{-2} = a^{-1}$

Now multiply all three factors, which adds the exponents. The common denominator is $6$.

$a^{\frac46 - \frac66 + \frac76} = a^{5/6}$

The radical is a power first, then the reciprocal flips the sign of the exponent.

$\frac{1}{\sqrt[4]{x^{3}}} = \frac{1}{x^{3/4}} = x^{-3/4}$

The index $4$ stays in the denominator of the exponent; writing $x^{-4/3}$ swaps the root and the power.

Distribute the outer exponent across the product, multiplying each inner exponent by $-\tfrac32$.

$\left(a^{2}b^{-4}\right)^{-3/2} = a^{2 \cdot \left(-\frac32\right)}\,b^{-4 \cdot \left(-\frac32\right)} = a^{-3}b^{6}$

Now send the negative exponent to the denominator.

$a^{-3}b^{6} = \frac{b^{6}}{a^{3}}$

Take each power separately, rooting before raising.

$81^{3/4} = \left(\sqrt[4]{81}\right)^{3} = 3^{3} = 27, \qquad 27^{2/3} = \left(\sqrt[3]{27}\right)^{2} = 3^{2} = 9$

Now subtract.

$27 - 9 = 18$

Dividing powers of one base subtracts the exponents, so check the quotient against the common denominator $12$.

$\frac{a^{5/6}}{a^{1/4}} = a^{\frac{10}{12} - \frac{3}{12}} = a^{7/12}$

The others miss: $a^{1/3} \cdot a^{1/4} \cdot a^{1/12} = a^{\frac{4}{12} + \frac{3}{12} + \frac{1}{12}} = a^{2/3}$, $\left(a^{7}\right)^{12} = a^{84}$, and $a^{1/3} + a^{1/4}$ is a sum of two powers, which never collapses into a single power.

The negative exponent turns the fraction upside down, which makes the exponent positive.

$\left(\frac{1}{32}\right)^{-3/5} = 32^{3/5} = \left(\sqrt[5]{32}\right)^{3} = 2^{3} = 8$

A base between $0$ and $1$ raised to a negative exponent gives an answer greater than $1$, never a negative one.

Evaluate the two rational powers of $64$ separately.

$64^{1/2} = \sqrt{64} = 8, \qquad 64^{1/3} = \sqrt[3]{64} = 4$

Now subtract the edge from the side.

$8 - 4 = 4$

The bigger denominator gives the smaller value: a higher root of a number greater than $1$ pulls it closer to $1$.